monte carlo methods and black scholes modelchristophe.chorro.fr/docs/malliavin1.pdf · monte carlo...

Monte Carlo Methods and Black Scholes model

Christophe Chorro ([email protected])

MASTER MMMEF

22 Janvier 2008

Christophe Chorro ([email protected]) (MASTER MMMEF)Monte Carlo Methods and Black Scholes model (some reminder) 22 Janvier 2008 1 / 87

Bibliography

The slides of this lecture (and other documents) are available on thefollowing website

http : //christophe.chorro.fr/enseignements.html

J. JACOD, P. PROTTER : Probability essentials, Springer, 2004.

Course of ANNIE MILLET in Paris 1: “ftp : //samos.univ −paris1.fr/pub/SAMOS/cours/millet/Master1/PolyM1_English.pdf ′′

http://christophe.chorro.fr/docs/CSangl.pdf

D. LAMBERTON, B. LAPEYRE : Introduction to stochastic calculus appliedto finance

MARK BROADIE AND PAUL GLASSERMAN, Estimating Security PriceDerivatives Using Simulation, Management Science, 1996, Vol. 42, No.2, 269–285.


Study Plan of the Lecture

Introduction to Monte Carlo MethodsPresentation of the problemAn historical exampleStrong law of large numbersControl of the error (the central limit theorem)Conclusion

Black scholes modelSimulations of Gaussian random variablesSimulation of the Brownian motionReminder on the Black Scholes modelThe greeksFinite difference method for GreeksIntegration by parts method for Greeks


Plan

1 Introduction to Monte Carlo MethodsPresentation of the problemAn historical exampleStrong law of large numbers (SLLN)Control of the error: the CLTConclusion

2 Black Scholes modelSimulations of Gaussian random variablesSimulation of Brownian motionReminder on the Black Scholes modelThe greeksFinite difference method for GreeksIntegration by parts method for GreeksNumerical resultsConcluding remarks


Presentation of the problem

The basic problem is to estimate a multi-dimensional integral∫Rd

f (x)dµ(x)

for which an analytic answer is not known.

More precisely we look for a stochastic algorithm that gives:

A numerical estimate of this integral,

An estimate of the error,

A good accuracy with an interesting computational cost.


Presentation of the problem

This kind of problem appears in a wide field of areas:

Von Neuman, Ulam: Neutron diffusion in fissionable material (Manhattanproject 1947)

Biology

Mathematical finance (pricing and hedging of contingents claims)


Plan




The Buffon needle

We consider a floor with equally spaced lines, a distance δ apart and a needleof length 0 < l < δ dropped on it

Question: What is the probability that a needle dropped randomly intersectsone of the lines?


The Buffon needle

If θ = θ0 is fixed,

P(Intersection) =l sinθ0

δ.

If the needle is dropped randomly, θ is uniformly distributed on [0, π[ and

P(Intersection) =

∫ π

0

l sinθ0

δ

dθ0

π=

2lπδ

.

Now, if we drop N needles and denote by X the number of them crossing aline, one has

XN≈ 2l

πδ.


The Buffon needle

Laplace (1812) suggested to use this experiment to approximate π:

π ≈ 2lNXδ

.

Lazzarini (1901) made the experiment with l = 2.5cm, δ = 3cm andN = 3408. He obtained X = 1808 thus

π ≈ 355133

≈ 2.66917....

Problem: Time consuming and bad accuracy.......


The Buffon needleSee http://www.mste.uiuc.edu/reese/buffon/bufjava.html for computersimulation of this experiment.


The Buffon needle in modern language

We consider a random variable Z such that

Z = 1 if the randomly dropped needle crosses a line

Z = 0 otherwise.

We have

E[Z ] = P(Intersection) =2lπδ

and if we denote by Z1, ......ZN a N-sample of Z we previously use the

following approximation

Z1 + ... + ZN

N≈ E[Z ].

Aim: Prove the validity of this approximation.


Plan




SLLN

We consider a probability space (Ω,A, P) and we denote by E the expectationunder P.

TheoremLet (Xn)n∈N be i.i.d random variables with values in R such that E[|X1|] < ∞.

Then, denoting Sn = X1 + ... + Xn, one has

Sn

n→

a.s and L1E[X1].


SLLN

Numerical illustration (Buffon needle)

0 50 100 150 200 250 300 350 400 450 5000.0

0.1

0.2

0.3

0.4

0.5

0.6

Illustration of the SLLN when X1 → B( 12 ) and n = 500


SLLN

Remark 1: The hypothesis E[|X1|] < ∞ is necessary

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000!25

!20

!15

!10

!5

0

5

10

SLLN is not fulfilled when X1 → C(1) (here n = 10000)

Remark 2: Possible extension for random variables with values in Rd


SLLN: Sketch of the proof

We suppose that E[|X1|4] < ∞ and without loss of generality we assume thatE[X1] = 0.

Using that the random variables are independent and centered we obtain

E[(

Snn

)4]

= 1n4

(n∑

k=1E[X 4

k ] + 3∑i 6=j

E[X 2i ]E[X 2

j ]

)= 1

n4

(nE[X 4

1 ] + 3n(n − 1)E[X 21 ]2)

≤CS

3E[X 41 ]

n2 .

Thus

E

[(Sn

n

)4]→ 0 ⇒ E

[∣∣∣∣Sn

n

∣∣∣∣]→ 0.


SLLN: Sketch of the proof

Moreover from the preceding inequality and the monotone convergencetheorem:

E

[ ∞∑n=1

(Sn

n

)4]

=∞∑

n=1

E

[(Sn

n

)4]

< ∞

thus∞∑

n=1

(Sn

n

)4

< ∞ a.s.

This implies that

Sn

n→a.s

0.


Plan




Control of the errorAn asymptotic bound via the Central Limit Theorem (CLT)

DefinitionWe say that a sequence of random variables (Xn)n>0 converges toward X indistribution ( Xn →

DX) if ∀f ∈ Cb(R, R),

E[f (Xn] →n→∞

E[f (X )].

This convergence extends when f is an indicator function.

TheoremLet (Xn)n∈N be i.i.d random variables with values in R such that E[|X1|2] < ∞.Then

Sn − nE[X1]√nσ

→DN (0, 1)

where σ2 = Var(X1).Christophe Chorro ([email protected]) (MASTER MMMEF)Monte Carlo Methods and Black Scholes model (some reminder) 22 Janvier 2008 20 / 87

Control of the error

Numerical illustration:

!! !" !# $ # " !$%$$

$%$&

$%#$

$%#&

$%"$

$%"&

$%!$

$%!&

$%'$

$%'&

Illustration of the CLT when X1 → U([0, 1]) and n = 500

We have an obvious extension of this result in any finite dimension



From the preceding theorem we obtain that

P(−aσ√

n≤ Sn

n− E[X1] ≤

aσ√n

)→

n→∞

∫ a

−a

1√2π

e−x22 dx .

In practice we know from tables (cf next slide that)

P(|N (0, 1)| ≤ 1.96) = 0.95

thus when n is large enough, with a confidence of 95%,

E[X1] ∈[

Sn

n− 1.96σ√

n,

Sn

n+

1.96σ√n

].

The magnitude of the error is given by 1.96σ√n : the size of σ is fundamental for

the speed of convergence.



When σ is unknown it may be estimated easily:

TheoremLet (Xn)n∈N be i.i.d random variables with values in R such that E[|X1|2] < ∞.

Then if we define

σn2 =

nn − 1

(1n

n∑i=1

X 2i − (

1n

n∑i=1

Xi)2

)one has

a)E[σn

2] = σ2

b)σn

2 →a.s

σ2.



We have the following result that allows the building of confidence intervalseven if σ is unknown.

TheoremLet (Xn)n∈N be i.i.d random variables with values in R such that E[|X1|2] < ∞,then

Sn − nE[X1]√nσn

→DN (0, 1)

Proof: It is just a consequence of the classical CLT and of the Slutsky lemma:

LemmaIf Xn →

DX and Yn →

Da (a being a constant) then

XnYn →D

Xa.


Plan




Reminder of the method

To basically compute E[f (X )] by Monte Carlo Methods we have to

“Generate” a n-sample of the distribution of X

Compute 1n

n∑k=1

f (Xk ) for large n

Precise the confidence interval coming from the CLT


Advantages

Easy to implement on any software

No regularity on f

The control of the error ( σ√n ) is independent of the dimension of the

problem


Limits

The error is a random variable (we only have confidence intervals)

This method may be slow if we don’t use extra-techniques


Limits

Consider that we want to approximate by Monte Carlo simulations

I = E[eβN (0,1)]

(even if the exact value is given by eβ22 ).

In this way we generate (see part) a n-sample (G1, ...., Gn) of a N (0, 1) anduse

In =eβG1 + ... + eβGn

n≈ E[eβN (0,1)].

By the CLT the order of magnitude of the relative error is given by

In − II

≈ σ√(n)I

=

√eβ2 − 1√

n.


Limits

When β = 1, we need n ≥ 1000 to obtain a relative error ≤ 10%

When β = 5, we need n ≥ 7 ∗ 1010 to obtain a relative error ≤ 100%(THIS IS NUMERICALLY IMPOSSIBLE) Moreover, in this case

Exact value= 268357

Approximated value (for n=100000)= 107709

Confidence interval (level 95%)= ]20188, 195229[


Plan




Simulations of Gaussian random variables

Starting from i.i.d uniform random variables on [0, 1] (Un)n∈N

( See: http://christophe.chorro.fr/docs/MC1.pdf)

we want to generate random samples of a Gaussian distribution.


Gaussian random variables from Inversion Method

Let X be a real random variable, the distribution function of X is defined as

FX (x) = P(X ≤ x); x ∈ R.

Properties: non-decreasing, right C0, limx→+∞

F (x) = 1 and limx→−∞

F (x) = 0.

DefinitionWe define the generalized inverse of FX denoted by F−X where ∀u ∈]0, 1[,

F−X (u) = infx | FX (x) ≥ u.

Remark: When FX is strictly increasing and continuous, F−X = F−1X .



PropositionIf U → U([0, 1]) then F−X (U) has the same distribution then X.

Proof: We just have to remark that ∀u ∈]0, 1[, ∀x ∈ R,

F−X (u) ≤ x ⇔ u ≤ FX (x).

Figure: Illustration of the definition of F−X



Example 1: Particular continuous distributions

If X → E(λ), λ > 0 then FX (x) = (1− e−λx)1x≥0 and

−1λ

Log(1− U) → E(λ).

If X → C(a), a > 0, then FX (x) = 1π [Arctan( x

a ) + π2 ] and

a tan(

π(U − 12

)

)→ C(a).



Example 2: When X → N (0, 1), FX is unknown but we have the followingapproximation:

If u > 0; 5, let t =√−2log(1− u)

F−X (u) ≈ t − c0 + t(c1 + tc2)

1 + t(d1 + t(d2 + td3)).

If u ≤ 0; 5, let t =√−2log(u)

F−X (u) ≈ c0 + t(c1 + tc2)

1 + t(d1 + t(d2 + td3))− t .

Where

c0 = 2.515517, c1 = 0.802853, c3 = 0.010328,

d1 = 1.432788, d2 = 0.189269, d3 = 0.001308.



−4 −2 0 2 4

0.0

0.1

0.2

0.3

0.4

Figure: Empirical density of 10000 standard normal distribution obtained by Inversionmethod


Gaussian random variables from Transformationmethod

Here we try to express X as a function of another random variable Y easy togenerate.

One of the main tool is the following result

PropositionLet D and ∆ be two open sets of Rd and Φ = (Φ1, ...Φd ) : D → ∆ aC1-diffeomorphism. If g : ∆ → R is measurable and bounded then∫

∆

g(v)dv =

∫D

g(Φ(u)) | JΦ(u)) | du

where JΦ(u) = det[(

∂Φi

∂uj(u))

1≤i,j≤d

].


Gaussian random variables from TransformationmethodExample 1: Box-Muller method

PropositionIf U1 and U2 are two independent uniform random variables on [0, 1] then

G1 =√−2log(U1)cos(2πU2) and G2 =

√−2log(U1)sin(2πU2)

are two independent N (0, 1).

Proof: Let us define the following C1-diffeomorphism

Ψ : (x , y) ∈]0, 1[2→ (u =√−2log(x)cos(2πy), v =

√−2log(x)sin(2πy))

fulfilling |JΨ(x , y)| = 2πx . Since u2 + v2 = −2log(x), according to the change

of variables theorem (Φ = Ψ−1), one has for F ∈ Cb(R2, R),∫]0,1[2

F (Ψ(x , y))dxdy =

∫R2−(R+×0)

F (u, v)1

2πe−

u2+v22 dudv .


Gaussian random variables fromTransformationmethod

−4 −2 0 2 4

−4

−2

02

4

Figure: Simulation of 5000 pairs of independent N (0, 1) by Box-Muller method


Plan




Simulation of Brownian motion

We consider a probability space (Ω,A, P).

DefinitionStandard Brownian motion (B.M) is a stochastic process (Bt)t∈[0,T ] fulfilling :

a) B0 = 0 P-a.s.

b) B is continuous i.e t → Bt(w) is continuous for P almost all w.

c) B has independent increments: For Si t > s, Bt − Bs is independent ofFBs = σ(Bu, u ≤ s).

d) the increments of B are stationary and gaussian: For t ≥ s, Bt − Bs followsa N (0, t − s).


Simulation of Brownian motion

We consider a subdivision 0 = t0 < ... < tn = T of [0, T ]. We want to simulate

(Bt0 , ...Btn).

Idea:Btk = Btk−1 + Btk − Btk−1︸︷︷︸

N (0,tk−tk−1)⊥Btk−1 ,...,B0

.

PropositionIf (G1, ...Gn) are i.i.d N (0, 1), we define

X0 = 0, Xi =i∑

j=1

√tj − tj−1Gj i > 0.

Then(X0, ..., Xn) =

D(Bt0 , ...Btn).


Brownian motion

0.0 0.4 0.8−

0.8

−0.

20.

40.0 0.4 0.8

−1.

5−

0.5

0.0 0.4 0.8

−0.

60.

00.

4

0.0 0.4 0.8−

0.5

0.5

Figure: 4 paths of the Brownian motion on [0, 1] generated using the precedingmethod with the regular subdivision of step 0.001.


Brownian motion

If we want to add points to the preceding simulation, the following result isuseful.

PropositionFor t > s,

D(

B t+s2| (Bt , Bs)

)= N

(Bt + Bs

2,

t − s4

).

Idea of the proof

B t+s2

=Bt + Bs

2+ Z

where

• Z is independent of σ(Bu | u ≥ t) and σ(Bu | u ≤ s).

• Z → N(0, t−s

4

).


Plan




Black Scholes model

We consider the time interval [0, T ] and r the risk free rate (supposed to beconstant) during this period.

Non-risky asset: Its dynamic is given by

S00 = 0, S0

t = ert .

Risky asset: Under the historical probability P its dynamic is given by thefollowing SDE:

dSt = µStdt + σStdBt (1)

with initial condition S0 = x0 > 0 and where B is a standard BM under P.

Itô formula ⇒ St = x0e(µ− 12 σ2)t+σBt .


Black Scholes model

0.0 0.4 0.81

35

sigma=1, mu=1

0.0 0.4 0.8

0.4

0.7

1.0

sigma=1, mu=−1

0.0 0.4 0.8

1.0

2.5

4.0

sigma=0.5, mu=1

0.0 0.4 0.80.

40.

71.

0

sigma=0.5, mu=−1

Figure: Simulation of a path of the risky asset in the B&S model for differentparameters


Black Scholes model

What is, in this model, the price at t of a contingent claim with payoff ΦT at T?

PropositionIn the B&S model there exists a unique probablity Q ∼ P such that the priceat t of a contingent claim with payoff ΦT at T is given by

Pt = E∗[e−r(T−t)ΦT | Ft ].

Moreover the dynamic of the risky asset under Q is given by

dSt = rStdt + σStdWt (µ ⇔ r) (2)

where W is a standard BM under Q.


Black Scholes model

Examples: The Black Scholes Formulas

For Call options (ΦT = (ST − K )+) one has

E∗[(ST − K)+ | Ft] = StN(d1(t, St))− Ke−r(T−t)N(d2(t, St))

where

d1(t , x) =log( x

K ) + (r + σ2

2 )(T − t)σ√

T − tet d2(t , x) =

log( xK ) + (r − σ2

2 )(T − t)σ√

T − t

and where N is the distribution function of a N (0, 1).

For Put options (ΦT = (K − ST )+) one has

E∗[(K− ST)+ | Ft] = −StN(−d1(t, St)) + Ke−r(T−t)N(−d2(t, St)).


Black Scholes modelThis formulas are fundamental because

They are easy to compute in practiceThey are a Benchmark to test numerical methodsIn the case where we don’t have closed form formulas we may use MonteCarlo Methods

price = e−rT E∗[ΦT ] ≈ e−rT 1N

N∑i=1

ΦiT

where the ΦiT are independent realizations of ΦT .

Morever we have a control of the error (CLT): with a probability of 95%

price ∈

[e−rT 1

N

N∑i=1

ΦiT −

1.96e−2rT Σ√N

, e−rT 1N

N∑i=1

ΦiT +

1.96e−2rT Σ√N

]

Σ being the (empirical) variance of ΦT .


Black Scholes model

Examples: We take x = 100, K = 100, σ = 0, 15, r = 0, 05, T = 1, t = 0.

Price of a call on mean: e−r E[

(S 1

2+S1

2 − K)

+

]

Estimated value (N=10000) = 6.05, confidence interval at 95% :[5.89; 6.20]Estimated value (N=100000) = 6.04, confidence interval at 95% : [5.99; 6.09]

Price of a call on max: e−r E[(Max(S 12, S1)− K )+] with σ = 0.5 and

K = 1.Estimated value (N=10000) = 8.68, confidence interval at 95% : [8.49; 8.87]Estimated value (N=10000) = 8.88, confidence interval at 95% : [8.82; 8.95]

For prices, the error only comes from Monte-Carlo approximations


Plan




The greeks

We suppose that the payoff of a contingent claim is given by ΦT = f (ST ), thus,

Pt = E∗[e−r(T−t)ΦT | Ft ] = F (t , St) (Markov process)

where

F (t , x) = e−r(T−t)∫ +∞

−∞f (xe(r− 1

2 σ2)(T−t)+σy√

T−t)1√2π

e−y2

2 dy .

Thus, F (t , x) = E∗[e−r(T−t)f (SxT−t)].

PropositionUnder mild hypotheses on f , F∈ C1,2([0, T [×R, R).


The greeks

The greeks measure the sensitivity of the price with respect to a givenparameter.

• ∆ measures the sensitivity of the price with respect to the underlying

∆t(St) =∂F∂x

(t , St)

It is also the quantity of risky asset in the hedging portfolio!!!!• Γ measures the sensitivity of the delta with respect to the underlying

Γt(St) =∂2F∂x2 (t , St)

It is also a measure of the frequence a position must be re-hedgedin order to maintain a delta neutral position


The greeks

• Θ measures the sensitivity of the price with respect to the underlying

Θt(St) =∂F∂t

(t , St)

• ρ measures the sensitivity of the price with respect to interest rate

ρt(St) =∂F∂r

(t , St)

• vega (which is not a greel letter!!!) measures the sensitivity of the pricewith respect to the volatility

vegat(St) =∂F∂σ

(t , St)

precautions to take for the estimation of σ!


The greeksFor call and put options ( strike K maturity T ) the greeks at t = 0 are given by:

Call Put

∆0(x) N(d1) > 0 −N(−d1) < 0

Γ0(x) 1xσ√

TN ′(d1) > 0 1

xσ√

TN ′(d1) > 0

Θ0(x) − xσ

2√

TN ′(d1)− Kre−rT N(d2) < 0 xσ

2√

TN ′(d1) + Kre−rT (N(d2)− 1) ??

ρ0(x) TKe−rT N(d2) > 0 TKe−rT (N(d2)− 1) < 0

vega0(x) x√

TN ′(d1) > 0 x√

TN ′(d1) > 0

where

d1(x) =log( x

K ) + (r + σ2

2 )T

σ√

Tet d2(x) =

log( xK ) + (r − σ2

2 )T

σ√

T(3)

and where N is the distribution function of a N (0, 1).Christophe Chorro ([email protected]) (MASTER MMMEF)Monte Carlo Methods and Black Scholes model (some reminder) 22 Janvier 2008 58 / 87

The greeks

• In the preceding table, to obtain the values of the greeks at time t we justhave to change T into T − t

• They are easy to compute in practice

• They are a Benchmark to test numerical methods

We will restrict ourselves to ∆ and Γ.


Plan




Finite difference method for Greeks

A classical method is to use a finite difference scheme

∆t(x) ≈ F (t , x + h)− F (t , x − h)

2h

Γt(x) ≈ F (t , x + h) + F (t , x − h)− 2F (t , x)

h2

where h is sufficiently small.

F (t , x + h), F (t , x − h) et F (t , x) are computed by Monte Carlo methods.

Reminder: F(t, x) = E∗[e−r(T−t)f(SxT−t)].



• Contrary to prices, for the greeks there are two factors of approximationtwo factors of approximation

Error from finite difference︸︷︷︸E1

+ Error from Monte Carlo︸︷︷︸E2

• The choice of h may be difficult (cf: Broadie-Glasserman):

• When h is too big, E1 may strongly increase.

• When h is too small, the variance of the Monte Carlo estimator may explode.

• When we use ∆t(x) ≈ F (t,x+h)−F (t,x−h)2h since

Var(F (t, x + h)− F (t, x − h)) = Var(F (t, x + h)) + Var(F (t, x + h))− 2Cov(F (t, x + h), F (t, x − h))

it is (often) better to use the same random sample for the two MonteCarlo simulations!



A priori (it will be confirmed by numerical results) this methods should bemore efficient when the prices are regular in x :

• When f (y) = 1y≥M (Binary options)

E∗[| f(Sx+hT )− f(Sx

T) |2] = P∗(

Mx + h

< e(r− 12 σ2)T+σBT <

Mx

)= O(h).

• When f (y) = (y − K )+ (Call)

E∗[| f(Sx+hT )− f(Sx

T) |2] ≤ E∗[(Sx+hT − Sx

T)2] = h2E∗[e2(r− 12 σ2)T+2σBT ] = O(h2).


Plan




Integration by parts method for Greeks

The idea is simple: Write the greeks on the following form

Greeks = E ∗ [PAYOFF× weight]

with

• A weight independent of the Payoff.

• A weight such that the variance of PAYOFF× weight is minimal.



Here we suppose that ΦT = f (ST ). The following result is independent of f :

PropositionOne has

∆t(x) = e−r(T−t)E∗[

BT−t

xσ(T − t)f (Sx

T−t)

]and

Γt(x) = e−r(T−t)E∗[(

−BT−t

x2σ(T − t)+

B2T−t − (T − t)(σ(T − t)x)2

)f (Sx

T−t)

].


Integration by parts method for GreeksProof in the case of delta: Let f ∈ C1

K (R, R) (use approximation otherwise).

According to the Lebesgue theorem of differentiation under the integral sign,

∆t(x) = e−r(T−t)∫ +∞

−∞

∂

∂xf (xe(r− 1

2 σ2)(T−t)+σy√

T−t)︸︷︷︸g(x,y)

1√2π

e−y2

2 dy .

But

∂g∂x

(x , y) =1

xσ√

T − t∂g∂y

(x , y).

Thus, using Integration by parts,

∆t(x) =e−r(T−t)

xσ√

T − t

∫ +∞

−∞f (xe(r− 1

2 σ2)(T−t)+σy√

T−t)y√2π

e−y2

2 dy

and

∆t(x) = e−r(T−t)E∗[

BT−t

xσ(T − t)f (Sx

T−t)

].



Advantages:

• One factor of approximation (Monte Carlo)

• This method doesn’t depend on the Payoff (the weight is independent off ).

Question:• Is there a criteria (in terms of variance) to choose among all the possible

weights?



PropositionLet Π being a square integrable weight such that for all Payoffs of the formf (ST )

Greek = E∗[f (ST )× Π].

Thus, the weight minimzing the variance of f (ST )× Π is given by

Π0 = E∗[Π | FT ].

Rk: The weights in the preceding proposion are optimal.



Proof: Let Π fulfilling greek = E∗[f (ST )Π], we want to minimize Var(f (ST )Π).

One has

Var(f (ST )Π) = E∗[(

f (ST )Π− greek)2]

= E∗[(

f (ST )(Π− Π0) + f (ST )Π0 − greek)2]

= E∗[(

f (ST )(Π− Π0))2]

+ Var(f (ST )Π0)

+ 2E∗[(

f (ST )(Π− Π0)(Π0f (ST )− greek))]

.

The last line being equal to zero, the result follows.


Plan




Numerical results Call-PutWe take x = 100, K = 100, σ = 0, 15, r = 0, 05, T = 1, t = 0.

Delta Call

0,64

0,645

0,65

0,655

0,66

0,665

0,67

1 5 9 13 17 21 25 29 33 37 41 45 49 53 57 61 65 69 73 77 81 85 89 93 97

Nbre de Simulations (10E4)

del

ta

Mall

DF

Valeur Theo:0,6584855


Numerical results Call-Put

Gamma Call

0,0225

0,023

0,0235

0,024

0,0245

0,025

0,0255

1 5 9 13 17 21 25 29 33 37 41 45 49 53 57 61 65 69 73 77 81 85 89 93 97

Nbre simulations(10E4)

Gam

ma Mall

DF

Valeur theo 0,0244688



Delta Put

-0,349

-0,346

-0,343

-0,34

-0,337

-0,334

1 5 9 13 17 21 25 29 33 37 41 45 49 53 57 61 65 69 73 77 81 85 89 93 97

Nbre simulations (10E4)

del

ta

Mall

DF

Valeur Theo -0,3415145



Gamma Put

0,0225

0,023

0,0235

0,024

0,0245

0,025

0,0255

0,026

1 5 9 13 17 21 25 29 33 37 41 45 49 53 57 61 65 69 73 77 81 85 89 93 97


gam

ma Mall

DF

Valeur Theo 0,0244688



Call Ratio of Variances (FD/ IBP)

∆ 0.134

Γ 0.132

Put Ratio of Variances (FD/ IBP)

∆ 0.30

Γ 1.32

• The payoff being regular, the finite difference performs quite well.• Integration by parts gives better results for Put than for Call (Explosion of

weight!!!)


Numerical results Digital

A Digital option is characterized by a payoff 1SxT >Kmin irregular in Kmin.

We can show easily that

F (0, x) = e−rT KN(d),

∆0(x) =e−rT

xσ√

Tn(d)

and

Γ0(x) =e−rT

x2σ2Tn(d)

(d + σ

√T)

where n is the density of a N (0, 1) and where d =log( x

Kmin)+(r−σ2

2 )(T )

σ√

T.


Numerical results DigitalWe take x = 100, K = 100, σ = 0, 15, r = 0, 05, T = 1, Kmin = 95.

Delta Digitale

0,02

0,0203

0,0206

0,0209

0,0212

0,0215

0,0218

0,0221

1 5 9 13 17 21 25 29 33 37 41 45 49 53 57 61 65 69 73 77 81 85 89 93 97


Del

ta

Mall

DF




Gamma Digitale

-0,0015

-0,0014

-0,0013

-0,0012

-0,0011

-0,001

-0,0009

-0,0008

-0,0007

-0,0006

-0,0005

1 5 9 13 17 21 25 29 33 37 41 45 49 53 57 61 65 69 73 77 81 85 89 93 97

Nbre Simulations (10E4)

Gam

ma Mall

DF




Digitale Ratio of Variances (FD/ IBP)

∆ 5.31

Γ 2354


Numerical results CorridorA corridor option is characterized by a payoff 1Kmax >Sx

T >Kmin (difference of twodigitals).

We take x = 100, K = 100, σ = 0, 15, r = 0, 05, T = 1, Kmin = 95,Kmax = 105.

Delta Corridor

-0,0052

-0,0047

-0,0042

-0,0037

-0,0032

1 5 9 13 17 21 25 29 33 37 41 45 49 53 57 61 65 69 73 77 81 85 89 93 97

Nbre de simulation (10E4)

Del

ta

Mall

DF

Valeur Théo: -0,0041146


Numerical results Corridor

Gamma Corridor

-0,002

-0,0015

-0,001

-0,0005

1 5 9 13 17 21 25 29 33 37 41 45 49 53 57 61 65 69 73 77 81 85 89 93 97

Nbre de simulations (10E4)

Gam

ma Mall

DF

Valeur Theo: -0,0009170


Numerical results Corridor

Corridor Ratio of Variances (FD/ IBP)

∆ 134

Γ 5785


Numerical results

• The more irregular is the payoff, the more efficient is the integration byparts method.

• The performance of the integration by parts method increases with theorder of derivation.

• In practice, the weights we found here are polynomials of WT . Thus theintegration by parts method will be more efficient for small maturities(small weights).

• The integration by parts method performs better for a put than for a call .

• The integration by part method is in fact a variance reduction technique..


Plan




Conclusions

For a payoff of the form f (St0 , ..., Stn) one has:

Proposition

∆0(x) = e−r(T−t)E∗[

f (St0 , ..., Stn)n∑1

λi(Bti − Bti−1)

]with λ1 = 1

xσt1et ∀1 ≤ i < n − 1

λi+1 =

1x −

i∑j=1

(tj − tj−1)λjσ

(ti+1 − ti)σ.

So we may use (with optimality) the integration by parts method in the BlackScholes model for discrete Lookback or asian options.


Conclusions

• What happens for other payoffs in the Black Scholes model???

• What happens for other models???


monte carlo methods and black scholes modelchristophe.chorro.fr/docs/malliavin1.pdf · monte carlo...

Documents